Amanita, let's solve the problem inside the green rectangle step by step.

Understanding the Problem:

• A ball is thrown vertically upward from the ground with a speed of 20 m/s.
• The height $x$ (in meters) after $t$ seconds is given by the formula:

$$x = 20t - 5t^2$$

• We are asked to find the maximum height the ball reaches, for $0 \leq t \leq 4$.

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Step-by-Step Solution Plan:

1. Recognize the Type of Equation:

   • The equation $x = 20t - 5t^2$ is a quadratic equation in $t$.
   • The graph of this equation is a parabola opening downwards (because the coefficient of $t^2$ is negative).

2. Find the Vertex of the Parabola:

   • The maximum value of $x$ (the height) will be at the vertex of the parabola.
   • The vertex for a quadratic equation $x = at^2 + bt + c$ occurs at $t = -\frac{b}{2a}$.
   • Here, $a = -5$, $b = 20$.

3. Calculate the Time at Which Maximum Height Occurs:

   • Substitute the values into the vertex formula to find $t$.

4. Check if the Time is Within the Allowed Range:

   • The problem states $0 \leq t \leq 4$. Make sure the time found is within this range.

5. Calculate the Maximum Height:

   • Substitute the value of $t$ back into the original equation to find the maximum height.

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Let's execute each step:

Step 1 & 2: Find the vertex (maximum point) of the parabola.

$t = -\frac{b}{2a} = -\frac{20}{2 \times (-5)}$

Let's calculate this value. Amanita, let's continue with the solution:

From the calculation, the time at which the ball reaches its maximum height is $t = 2$ seconds.

Now, let's find the maximum height by substituting $t = 2$ into the original equation:

$$x = 20t - 5t^2$$ $$x = 20 \times 2 - 5 \times 2^2$$

After calculating, we find:

$$x = 40 - 20 = 20$$

Final Answer:
The highest point the ball reaches is 20 meters.